Question

Solve the following equations:
$\frac{x^3+1}{x^2-1}=x+\sqrt{\frac{6}{x}}$.

Answer 1

$\frac{x^3+1}{x^2-1}=x+\sqrt{\frac{6}{x}}\frac{(x+1)(x^2+1-x)}{(x+1)(x-1)}=x+\sqrt{\frac{6}{x}}\frac{(x^2+1-x)}{x-1}=x+\sqrt{\frac{6}{x}}$

where $(x\ne 1)$

$\frac{(x^2+1-x)}{x-1}-x=\sqrt{\frac{6}{x}}\frac{x^2+1-x-x^2+x}{x-1}=\sqrt{\frac{6}{x}}\frac{1}{x-1}=\sqrt{\frac{6}{x}}\sqrt{x}=\sqrt{6}(x-1)$

Squaring both sides, we get

$x=6{(x-1)}^{2}x=6x^2+6-12x6x^2+6-13x=06x^2-9x-4x+6=03x(2x-3)-2(2x-3)=0(3x-2)(2x-3)=0\Rightarrow x=\frac{2}{3},\frac{3}{2}$